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  • Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise  Very short answer question 39 math

Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise  Very short answer question 39 math

Answers (1)

Answer: \frac{-\pi}{6}

Given:

\sin ^{-1}\left(-\frac{1}{2}\right)

Hint:

\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)  is the range of the principal value branch of inverse sine function.

Solution:

\begin{aligned} \text { Let } y=& \sin ^{-1}\left(-\frac{1}{2}\right) \\\\ & \sin y=-\frac{1}{2}=\sin \left(-\frac{\pi}{6}\right) \\\\ & y=-\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \end{aligned}

Here,

\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]  is range of principal value of sin

\sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}

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